K.C.S.E Biology Q & A - MODEL 1999PP1QN16
Describe the:
a) Process of inhalation in mammals. b) Mechanisms of opening and closing of stomata in plants.
answers
a) Muscles of diaphragm contract; causing the diaphragm to flatten (from dome position. The external intercostals muscles contract internal intercostals muscles relax pulling the ribcage upward/forward and outward in man. These movements increases the volume of the thoracic cavity; reducing the pressure; of the thoracic cavity; compared to atmospheric pressure; this causes the atmospheric air to rush into the lungs. (Through the nostrils, trachea bronchioles and alveoli).
b) Theory- photosynthesis Guard cells have chloroplasts; in the presence of light; photosynthesis occurs in guard cells, producing sugar in guard cells; osmotic pressure increases/osmotic potential lowers; water from neighboring /adjacent cells enter into guard cells; causing turgidity of guard cells; causing turgidity of guard cells. Theory 1. Guard cells have chloplasts; in the presence of light photosynthesis occur in the guard cells of stomata; producing in the guard cells; osmotic pressure increases/lowers osmotic potential water from the neighboring /adjacent cells, enter into guard cells; causing turgidity of guard cells The inner walls of the guard cells are thicker than outer walls; so during turgidity the inner walls stretch more; causing the guard cells to bulge outward; stomata opens. Theory 2. Guard cells have chloroplasts (Day) in light; photosynthesis occurs in the leaf/guard cells lowering the CO2 concentrations; this increases PH/alkalinity which triggers of enzymatic conversion of starch to sugar (glucose); leading to low osmotic potential/ increased osmotic pressure in guard cells; guard cells absorb water from epidermal cells; thus becoming turgid; the inner walls are thicker than the outer walls; outer walls stench more than inner walls; causing guard cells to bulge outwards, stomata opens; In the absence of light (night); no photosynthesis; CO2 concentration increases due to respiration; PH lowered/ acidity increases; sugar converted to starch; osmotic pressure lowered/ osmotic potential increases; guard cells lose water to adjacent epidermal cell becoming flaccid; stomata close. Day low H+ high PH opens stomata. Starch glucose. Theory 3 Guard cells have chloroplasts; in light AT produced; the energy drives K+ irons from adjacent epidermal cells into guard cells; accumulation of K+ raises osmotic pressure (lower osmotic potential) of guard cells; guard cells absorbs water from adjacent epidermal cells; becoming turgid; the inner walls are thicker than the outer walls so outer walls stretch more than inner walls causing guard cells to bulge outward. Stomata opens. In the absence of light (night ) ATP rapidly decreases; no energy of potassium +ions pump ion; migrate by diffusion from guard cells to adjacent epidermal cells; become flaccid; the thinner outer walls of guard cells shrink
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K.C.S.E Biology Q & A - MODEL 1999PP1QN15
An experiment was carried out to investigate haemolysis of human red blood cells. The red blood cells were placed in different concentrations of sodium chloride solution. The percentage of haemolysed cells was determined. The results were as shown in the table below.
a) i) On the grid provided, plot a graph of harmolysed red blood cells against salt concentration.
ii) at what concentration of salt solution was the proportion of haemolysed cells equal to non-haemolysed cells? iii) State the percentage of cells haemolysed at salt concentration of 0.45% b) Account for the results obtained at: i) 0.33 percent salt concentration. ii) 0.48 percent salt concentration. c) What would happen to the red blood cells if they were placed in 0.50 percent salt solution? d) Explain what would happen to onion epidermal cells if they were placed in distilled water.
answers
a) (i)0.403; 0.404; + 0.002
ii) 0.402; iii) 9-10-11% b) Account for the results obtained at: (i) 0.33 percent salt concentration. Less concentration /hypotonic / dilute than blood cells cytoplasm/ red blood cells; water is drawn in by osmosis the cells swells and eventually burst. (ii)0.48 (ii)0.48 percent salt concentration. Concentration of cytoplasm same as concentration of salt solution/isotonic; therefore no net movement of water; hence no heomolysis. c)Percentage of cells haemolysed would still be zero? Becomes turgid; but does not burst; due to the cell wall. d)The cells would absorb water due to osmosis, swell and become turgid. The cell sap move conc. than surrounding water gate into the cell by osmosis; the cell swells/becomes turgid; but does not burst due to the cell wall
K.C.S.E Biology Q & A - MODEL 1998PP1QN20
Explain how the mammalian intestines are adapted to perform their function.
answers
​The mammalian intestines are relatively long/coiled/folded. This allows food enough (enough) lime/increases surface area for digestion and absorption of products of digestion. The intestinal surface area for absorption. The glands have enzymes which secrete enzymes for digesting e.g. of correct enzyme, maltase, sucrase, lactase, enterokinase and peptidases. Some glands/goblets cells also produce mucus which protects the intestinal wall from being digested, reduce friction. Intestines have opening of ducts which allows bile/pancreatic juice into the lumen. The intestines have circular and longitudinal muscles whose contraction/relaxation/peristalis leads to the mixing of food with acc. At least enzymes/juices facilitating rapid digestion and helps pus food along the gut. Intestines are well supplied with blood vessels to supply oxygen/remove digested food. Presence of lacteal vessels for transport of fats/lipids.
Have thin epithelia to facilitate fast/rapid absorption/diffusion. Allow increase in surface area for absorption only. Cell biology/cytology. Occurrence of cell e.g. mitochondria, ribosome’s, nucleas, cytochromes organelle point to a common ancestry.
K.C.S.E Biology Q & A - MODEL 1998PP1QN19
Discuss the various evidences, which show that evolution has taken place.
answers
​Comparative anatomy/taxonomy.
K.C.S.E Biology Q & A - MODEL 1997PP1QN20b
Describe how the mammalian small intestine is adapted to its function
answers
K.C.S.E Biology Q & A - MODEL 1997PP1QN20.
What is meant by the term digestion?
answers
K.C.S.E Biology Q & A - MODEL 1997PP1QN19
Describe how the tapeworm is adapted to a parasitic mode of life
answers
Has hooks/suckers: for attachment to wall of intestines: long; to increase surface area for absorption of food: award increase in S.A for absorption once. Secretes enzymes/to neutralize digestive enzymes; (mucus inhibitor substance/anti enzymes) Hermaphroditic: to ensure reproductive/ self fertilization. Production of many eggs: to ensure survival
Segment for egg dispersal: More than one host; for transmission: e.g T solium – pig (Intermediate host) T. Saginata. Long to fit in the intestine/ increase surface area for ( flatten) Absorption of food; Anaerobic survive in the gut with low Oxygen
K.C.S.E Biology Q & A - MODEL 1997PP1QN19
What is parasitism?
answers
K.C.S.E Biology Q & A - MODEL 1997PP1QN18
An experiment was carried out to determine the growth rates of bamboo and a variety of maize plants in two adjacent plots. The average height and average dry weight of plants from the two populations were determined over a period of twenty weeks. The data is as shown in the table below.
(a) Between which two weeks did the greatest increase in weight occur in
(b) Bamboo plants (ii) Maize plants (b) (i) Which of the two types of plants had a higher productivity by the end of the experiment (ii) Give a reason for your answer in (b) (i) above (c) Between weeks 14 and 18, the average height of the maize plants remained constant while average dry weight increased. Explain this observation (d) Suggest how the change in the average dry weight bamboo and maize Plants would have been at week 22 if the experiment was continued. (e) Why was it appropriate for this experiment to use (i) Dry weight instead of fresh weight (ii) Weight and height (f) Describe how the average height and weight of the plants were determined in this experiment. Average height Average dry Weight (g) Give a reason why secondary thickening does not occur in bamboo and maize plants
answers
(a) (i) Bamboo plants
4 and 6 (ii)Maize plants 12 and 14 (b) (i) Bamboo (ii) It had accumulated more weight and therefore greater dry weight (c) Maize plants have reached maturity/maximum height food being manufactured (in green parts); is utilized for growth storage primary in the cob. (d)Increase in weight – bamboo reject both increase/ decrease accept bamboo and maize increase/ decrease. (e) (i) Dry weight instead of fresh weight Fresh weight is dependant on the amount of water present in the plants and this fluctuates depending on environmental factors. (ii) Weight and height Both given a better measure of growth (f) Average height At every 2 weeks measure the height of samples of plants in each plot: Divide the total height by the number of plants in each of plot. Average dry weight Harvest the sample measure of the plants in each plot; dry to constant weight: And divide by the number of plants (g) Being monocots/ lack (Inter) fascicular cambium:
K.C.S.E Biology Q & A - MODEL 1996PP1QN 22
Describe how new plants arise by asexual reproduction
ANSWERS
K.C.S.E Biology Q & A - MODEL 1996PP1QN 21
Explain how the mammalian skin is adapted to perform its functions
ANSWERS
​The cornified layer is made up of dead cells, that prevent entry of bacteria and prevent physical damage; melanin protects the body against U-V variation; sebaceous glands produce a chemical/ ring substance which is of blood vessel; which when the body temperature is high dilate and heat is lost or when body temp is low blood vessels constrict.
And heat is retained. Hair when it is called, stands and traps air between themselves; to retain heat/ stop heat loss or when it is hot hair lies flat close on the skin; so does not trap air, and therefore heat is retained and sweat is lost; the skin has sweat glands which produces sweat; sweat evaporates thus cooling the body. K.C.S.E Biology Q & A - MODEL 1996PP1QN 20
A culture of bacteria was incubated in nutrient agar at 350C. Samples were taken at intervals in order to estimate the number of bacteria in the population. The data obtained is shown in the graph below.
​(a) When was the pollution of bacteria 350 million
(b) Account for the shape of the graph between (i) A and B (ii) B and C (iii) C and D (c) Give three reasons for the shape of the curve between D and E (d) (i) Suggest what would happen to the population of the bacteria if the temperature was lowered to 00 after incubating for 12 hours. (ii) Give a reason for your answer in (d) (i) above (e) Give three reasons why it is important to control human population growth rate in Kenya?
ANSWERS
​(a) 10 HRC and 31 HRC
(b) (i) A and B The number of bacteria dividing are few: bacteria are adjusting conditions: few are dying therefore high increase in population (ii) B and C More cells are dividing due to suitable environment/ favorable conditions; few are dying; therefore high increase in population (iii) C and D No population change; number produced is equal to number dying. (c) Accumulation of toxic wastes; that kills bacteria; depletion of nutrients leading to competition of space. (d) (i) The population will remain the same (ii) Temperature not conclusive for division (e) Food to be sufficient for population Social amenities/ education; health services (a) Describe how insect pollinated flowers are adapted to pollination |
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